HackerEarth Distinct Count problem solution

In this HackerEarth Distinct Count problem solution we have given an array A of N integers, classify it as being Good Bad, or Average. It is called Good, if it contains exactly X distinct integers, Bad if it contains less than X distinct integers, and Average if it contains more than X distinct integers.

HackerEarth Distinct Count problem solution.

#include <bits/stdc++.h>
using namespace std;
int main () {
    int tc;
    scanf("%d",&tc);
    while (tc--) {
        int n, k;
        scanf("%d%d",&n,&k);
        set < int > s;
        int temp;
        for (int i=0; i<n; i++) {
            scanf("%d",&temp);
            s.insert(temp);
        }

        if (s.size()<k) {
            printf("Bad husband\n");
        }
        if (s.size()==k) {
            printf("Perfect husband\n");
        }
        if (s.size()>k) {
            printf("Lame husband\n");
        }
    }
    return 0;
}

Second solution

#include<bits/stdc++.h>
using namespace std;
#define ll              long long int
#define vi              vector<int>
#define vl              vector<ll>
#define pii             pair<int,int>
#define pil             pair<int, ll>
#define pll             pair<ll, ll>
#define pli             pair<ll, int>
#define pb(v, a)        v.push_back(a)
#define mp(a, b)        make_pair(a, b)
#define MOD             1000000007
#define rep(i, a, b)    for(i=a; i<=b; ++i)
#define rrep(i, a, b)   for(i=a; i>=b; --i)
#define si(a)           scanf("%d", &a)
#define sl(a)           scanf("%lld", &a)
#define pi(a)           printf("%d", a)
#define pl(a)           printf("%lld", a)
#define pn              printf("\n")
ll pow_mod(ll a, ll b)
{
    ll res = 1;
    while(b)
    {
        if(b & 1)
            res = (res * a) % MOD;
        a = (a * a) % MOD;
        b >>= 1;
    }
    return res;
}
set<int> s;
int main()
{
    int t, i, n, x, j, tmp;
    si(t);
    assert(t >= 1 && t <= 50);
    rep(i, 1, t)
    {
        s.clear();
        si(n);
        si(x);
        assert(n >= 1 && n <= 13000);
        assert(x >= 1 && x <= 13000);
        rep(j, 1, n)
        {
            si(tmp);
            assert(tmp >= 1 && tmp <= 1000000000);
            s.insert(tmp);
        }
        if(s.size() < x)
            printf("Bad husband\n");
        else if(s.size() > x)
            printf("Lame husband\n");
        else
            printf("Perfect husband\n");
    }
    return 0;
}